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We will be working over an algebraically closed field of characteristic 0. We say that a projective variety $X\subset \mathbb{P}^n$ has projectively isomorphic plane sections if there is an open set $U\in(\mathbb{P}^n)^\vee$ such that the hyperplane sections $H\cap X,\ H\in U$ are all projectively isomorphic, i.e. for any two $H_1,H_2\in U$ there is some automorphism $f$ of $\mathbb{P}^n$ such that $f(H_1)=H_2, f(X\cap H_1)=X\cap H_2$.

Consider a smooth cubic hypersurface $X\subset \mathbb{P}^3.$ According to Theorem 2.12 of the paper "Some Remarks on Varieties with Projectively Isomorphic Hyperplane Sections" by Rita Pardini, $X$ has projectively isomorphic hyperplane sections if and only if it's a projection of a normal rational ruled surface in $\mathbb{P}^4.$ I suspect that this is not the case in this situation.

Is there a way to see that $X$ does not have projectively isomorphic hyperplane sections? More generally, what are the ways to identify such projections of normal rational ruled surfaces of degree $d$ from $\mathbb{P}^{d+1}$ to $\mathbb{P}^d$?

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    $\begingroup$ Note that such a (smooth) surface exists in characteristic 2, namely the Fermat cubic $X^3+Y^3+Z^3+T^3=0$. $\endgroup$
    – abx
    Sep 26, 2021 at 13:34

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Try this: Consider a "Lefschetz pencil" of planes through a general line. I believe these cut the smooth cubic surface in a pencil of irreducible plane cubic curves that are generally smooth, and the special ones have only one node, in particular are stable curves. Hence the smooth ones converge to the nodal ones in the moduli space of stable curves of genus one. That space is hausdorff, so the smooth ones cannot all be isomorphic, much less projectively equivalent. Does this work?

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  • $\begingroup$ Thanks! This seems to work! $\endgroup$ Sep 26, 2021 at 12:09
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According to Corollary 1.5 of that paper, if $X$ has projectively isomorphic hyperplane sections, then the irreducible ones are uniruled. In your case this does not happen because the general section is an irreducible plane cubic which is not a rational curve.

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  • $\begingroup$ Oh wow, I can't believe I missed that... Thank you so much! $\endgroup$ Sep 25, 2021 at 17:00
  • $\begingroup$ Actually, I looked over corollary 1.5 again and now think this is incorrect. Just realized that I, and maybe you too, once again got confused by the notation in the paper. In Corollary 1.5, $Y$ is a special hyperplane section, not the generic one. It's the special sections that must be uniruled, not the generic ones. The generic section is uniruled for a variety of dimension 3 or more, according to Proposition 1.7. $\endgroup$ Sep 26, 2021 at 5:18

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